Two square metal plates, 6.0 cm on a side, face each other in air at a distance of 0.40 mm…

Step 1. The capacitance. $A = (0.060\ \text{m})^2 = 3.6 \times 10^{-3}\ \text{m}^2$. $C = \dfrac{\varepsilon_0 A}{d} = \dfrac{8.85 \times 10^{-12} \times 3.6 \times 10^{-3}}{0.40 \times 10^{-3}} = 7.97 \times 10^{-11}\ \text{F}$, that is $79.7$ pF.

Step 2. The charge. $Q = CV = 7.97 \times 10^{-11}\ \text{F} \times 9.0\ \text{V} = 7.17 \times 10^{-10}\ \text{C}$, that is $0.717$ nC.

Step 3. The field. $E = \dfrac{V}{d} = \dfrac{9.0\ \text{V}}{0.40 \times 10^{-3}\ \text{m}} = 2.25 \times 10^{4}\ \text{V/m}$, that is $22.5$ kV/m.

Step 4. The charge on each square metre. $\sigma = \dfrac{Q}{A} = \dfrac{7.17 \times 10^{-10}\ \text{C}}{3.6 \times 10^{-3}\ \text{m}^2} = 1.99 \times 10^{-7}\ \text{C/m}^2$, that is $0.199\ \mu\text{C/m}^2$.